arXiv:1310.0897v4 [math.NT] 4 Mar 2014
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
TIANXIN CAI, DEYI CHEN AND YONG ZHANG
Abstract. In this paper, we consider some hybrid Diophantine equations of
addition and multiplication. We first improve a result on new Hilbert-Waring
problem. Then we consider the equation
(
A+B =C
(1)
ABC = D n
where A, B, C, D, n ∈ Z+ and n ≥ 3, which may be regarded as a generalization
of Fermat’s equation xn +y n = z n . When gcd(A, B, C) = 1, (1) is equivalent to
Fermat’s equation, which means it has no positive integer solutions. We discuss
several cases for gcd(A, B, C) = pk where p is an odd prime. In particular,
for k = 1 we prove that (1) has no nonzero integer solutions when n = 3 and
we conjecture that it is also true √
for any prime n > 3. Finally, we consider
equation (1) in quadratic fields Q( t) for n = 3.
1. Introduction
In this paper, we consider some hybrid Diophantine equations of addition and
multiplication. First of all,
such that
n = x1 + x2 + · · · + xs
x1 x2 · · · xs = xk ,
for n, xi , x, k ∈ Z+ , which is a new variant of Waring’s problem:
n = xk1 + xk2 + · · · + xks .
We denote by g ′ (k) (resp. G′ (k)) the least positive integer such that every integer
(resp. all sufficiently large integer) can be represented as a sum of at most g ′ (k)
(resp. G′ (k)) positive integers, and the product of the g ′ (k) (resp. G′ (k)) integers
is a k-th power. We show [2] that
g ′ (k) = 2k − 1; G′ (p) ≤ p + 1; G′ (2p) ≤ 2p + 2 (p ≥ 3); G′ (4p) ≤ 4p + 2 (p ≥ 7);
where k is a positive integer and p is prime. In this paper, we improve the results
on composite numbers as follow.
Theorem 1. For any composite number k, G′ (k) ≤ k + 2.
Next, we consider Fermat’s Last Theorem. In 1637, Fermat claimed that the
Diophantine equation
xn + y n = z n
has no positive integer solutions for any integer n ≥ 3. This was proved finally by
Andrew Wiles in 1995 [11, 13].
There are several generalizations of Fermat’s Last Theorem, e.g., Fermat-Catalan
conjecture, which states that the equation am + bn = ck has only finitely many
2010 Mathematics Subject Classification. Primary 11D41; Secondary 11D72.
Project supported by the National Natural Science Foundation of China 11351002.
1
2
TIANXIN CAI, DEYI CHEN AND YONG ZHANG
solutions (a, b, c, m, n, k) , where a, b, c are positive coprime integers and m, n, k are
1
+ n1 + k1 < 1. So far there are only 10 solutions found
positive integers, satisfying m
[4, 10]. Meanwhile, Beal’s conjecture [8] states that the equation Ax + B y = C z
has no solution in positive integers A, B, C, x, y and z with x, y and z at least 3
and A, B and C coprime. Beal has offered a prize of one million dollars for a
proof of his conjecture or a counterexample [14]. Obviously, there are only finite
solutions for Beal’s equation under Fermat-Catalan conjecture. Meanwhile, it’s
known that both FLT and Fermat-Catalan conjecture are the consequences of the
abc-conjecture, the latter was claimed to be proved in 2012 but not confirmed yet
by Japanese mathematician Shinichi Mochizuki [9].
Now we expand the same idea to Fermat’s equation as we did before to HilbertWaring problem. We consider a new Diophantine equation
(
A+B =C
(1)
ABC = Dn
where A, B, C, D, n ∈ Z+ and n ≥ 3.
It is easy to see that if gcd(A, B, C) = 1 then A, B, C are pairwise coprime. Therefore,
Equation (1) has no positive integer solutions for gcd(A, B, C) = 1 ⇐⇒ FLT
(2)
In view of (2), we may ask some natural questions concerning (1) :
1. Is it possible to have a solution for all n ≥ 3? If so, is it possible to have infinitely
many solutions ?
2. Is it possible to have a solution when gcd(A, B, C) = pk , where p is a prime,
k ∈ Z+ ?
In this paper, we answer the first question affirmatively by proving the following
result.
Theorem 2. For any n 6≡ 0 (mod 3), n ≥ 3, (1) has infinitely many positive integer
solutions; for any n ≡ 0 (mod 3), n ≥ 3, (1) has no positive integer solutions.
For the second question, we discuss the special cases n = 4, 5 and obtain the
following
Theorem 3. If gcd(A, B, C) = pk where k ∈ Z+ , p is odd prime and p ≡ 3
(mod 8), then the equation
(
A+B =C
(3)
ABC = D4
has no positive integer solutions.
For p = 2 and some p ≡ 1, 5, 7 (mod 8), it is possible for (3) to have a positive
integer solution when gcd(A, B, C) = p. For example
(
(
(
17 + 272 = 289
5 + 400 = 405
2+2=4
4
4
5 × 400 × 405 = 304 ,
2×2×4=2 ,
17 × 272 × 289 = 34 ,
(
47927607119 + 1631432881 = 49559040000
47927607119 × 1631432881 × 49559040000 = 443679604,
where p = gcd(A, B, C) = 2, 17, 5, 239, so p ≡ 1, 5, 7 (mod 8) respectively.
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
3
Theorem 4. If gcd(A, B, C) = pk where k ∈ Z+ , p is odd prime and p ≡
6 1
(mod 10), then the equation
(
A+B =C
(4)
ABC = D5
has no positive integer solutions.
In general, we have the follows:
Conjecture 1. If n ≥ 3 is prime, gcd(A, B, C) = pk where k ∈ Z+ , p is odd prime
and p 6≡ 1 (mod 2n), then (1) has no positive integer solutions.
Finally, if n > 3 is prime, we construct special prime p such that (1) has positive
integer solutions for gcd(A, B, C) = pk as following
Theorem 5. If n > 3 is prime, n ≡ r (mod 3), 1 ≤ r ≤ 2, a, b, m ∈ Z, m 6= 0,
n
+bn
= p is an odd prime and a + b = mn , then p ≡ 1 (mod 2n) and
such that a a+b
(mod n),
(1) has positive integer solutions for gcd(A, B, C) = pk where k ≡ rn−1
3
k ∈ Z+ .
Let a = 2, b = −1, m = 1, we obtain
Corollary 1. If n > 3 is prime, n ≡ r (mod 3), 1 ≤ r ≤ 2 and p = 2n − 1
is Mersenne prime, then (1) has positive integer solutions for gcd(A, B, C) = pk
(mod n), k ∈ Z+ .
where k ≡ rn−1
3
Moreover, we have the following
Conjecture 2. If n > 3 is prime, n ≡ r (mod 3), 1 ≤ r ≤ 2 and gcd(A, B, C) =
pk where p is prime and k 6≡ rn−1
(mod n) , then (1) has no positive integer
3
solutions.
In particular, if n > 3 is prime, then 4 6≡ rn (mod n), so 1 6≡
follows that we have a special case of Conjecture 2 when k = 1:
rn−1
3
(mod n), it
Conjecture 3. If n is odd prime, gcd(A, B, C) = p is prime, then (1) has no
positive integer solutions.
Remark. If n = 3, Conjecture 3 is true by Theorem 2. If abc-conjecture is true,
then Conjecture 3 should be true for fixed prime p and sufficiently large n where n
need not be prime.
B C
Proof. Because gcd(A, B, C) = p, then gcd( A
p , p , p ) = 1. By (1) we have
(
B
C
A
p + p = p
ABC = Dn .
n
So rad ABC
= rad pD3 ≤ rad(D) and C > D 3 . For any n ≥ 7 and 0 < ǫ < 31 ,
p3
we deduce
1
7
7
p ≤ D = D 3 −1− 3 ≤ D 3 −1−ǫ
and
n
log ( Cp )
log D − ( 73 − 1 − ǫ) log D
A B C
≥ 3
=
≥ 1 + ǫ.
q − ,− ,
p
p p
log D
log rad( ABC
p3 )
4
TIANXIN CAI, DEYI CHEN AND YONG ZHANG
B C
By abc-conjecture, there exist only finitely many triples − A
p , − p , p . Let A1 =
= Bp , C1 = Cp , then gcd(A1 , B1 , C1 ) = 1, p3 A1 B1 C1 = Dn . Let M be the
greatest m such that there is prime q satisfying q m |A1 B1 C1 . Then, if n > M + 3,
there is no solution for p3 A1 B1 C1 = Dn . Hence, when gcd(A, B, C) = p, (1) has
no positive integer solutions for sufficiently large n.
A
p , B1
However, we could not deduce Conjectures 1-3 from abc-conjecture.
2. Preliminaries
Lemma 1. [3, Proposition 6.5.6] Let c be a nonzero integer. The equation x4 −y 4 =
cz 2 has a rational solution with xyz 6= 0 if and only if |c| is a congruent number.
Lemma 2. [12, Proposition 5] Let p be a prime congruent to 3 modulo 8, then p
is not a congruent number.
Lemma 3. [5] Let A > 2 be a positive integer and has no prime divisors of the
form 10k + 1, then the equation
(
x5 + y 5 = Az 5
(5)
gcd(x, y) = 1
has no nonzero integer solution. If A = 2, the solutions of (5) are (x, y, z) =
±(1, 1, 1).
This lemma was first conjectured by V. A. Lebesgue [7] in 1843 and proved by
E. Halberstadt and A. Kraus [5] in 2004.
Lemma 4. For any prime p, integer n ≥ 2, if gcd(A, B, C) = pk and k ≡ 0
(mod n), k ∈ Z+ , then (1) has no nonzero integer solutions.
Proof. Let A1 = pAk , B1 = pBk , C1 = pCk , in view of gcd(A, B, C) = pk and (1), we
obtain that A1 , B1 , C1 are pairwise coprime and (1) can be changed into
(
A1 + B1 = C1
p3k A1 B1 C1 = Dn .
But k ≡ 0 (mod n), A1 , B1 , C1 are pairwise coprime, so A1 = xn , B1 = y n ,
C1 = z n and xn + y n = z n . By Fermat’s Last Theorem, we deduce that xyz = 0,
so ABC = 0. Contradiction.
3. Proofs of the Theorems
Proof of Theorem 1. For every positive integer n, let n = km + r where 0 ≤ r ≤
k − 1.
If r = 0, when n > 2k 2k , we have m = nk > 2k 2k−1 and
n = km = (m − 2k 2k−1 ) + · · · + (m − 2k 2k−1 ) +k 2k + k 2k .
{z
}
|
k
If 0 < r ≤ k − 1, when n > k
2k−1
+ k, we have m =
n−r
k2k−1
> k k−1 rk−1
k >
k
k k−1
k−1 k−1
n = km + r = (m − k k−1 rk−1 ) + · · · + (m − k
{z
|
k
r
) +k r
}
and
+ r.
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
5
Proof of Theorem 2. If n 6≡ 0 (mod 3), then there exists k ∈ Z+ such that
3k + 2 ≡ 0 (mod n). It is well known that there exist infinitely many (a, b, c) ∈ Z3+
such that a2 + b2 = c2 . Let

k+2 k k

b c
A = a
k k+2 k
B=a b
c


C = ak bk ck+2 .
So we have infinitely many positive solutions (A, B, C) satisfying (1), where D =
3k+2
(abc) n .
If n ≡ 0 (mod 3), suppose (A, B, C) is a positive solution of (1). Let d =
B
C
3
n
gcd(A, B, C). Then A
d , d and d are pairwise coprime and d |D . But 3|n, so
n
we have d|D 3 and
(
A
B
C
d + d = d
A
d
·
B
d
·
C
d
n
= ( Dd3 )3 .
B
C
A
3 B
3 C
3
3
Since A
d , d and d are pairwise coprime, so d = x , d = y , d = z and x +
3
3
y = z . By Fermat’s Last Theorem, we deduce that xyz = 0 and ABC = 0.
Contradiction.
Proof of Theorem 3. In view of Lemma 4, we only need to discuss k 6≡ 0
(mod 4). Suppose odd prime p ≡ 3 (mod 8) and (3) has a positive integer solution (A, B, C). In view of gcd(A, B, C) = pk , (3) can be changed into
(
A1 + B1 = C1
(6)
p3k A1 B1 C1 = D4 .
where A1 = pAk , B1 = pBk and C1 = pCk are pairwise coprime. But k 6≡ 0 (mod 4),
we deduce that only one of A1 , B1 , C1 is congruent to 0 modulo p. Let 3k ≡ r
(mod 4), then 1 ≤ r ≤ 3. In view of (6) and reorder A1 , B1 if necessary, we obtain

4

A1 = x
(7)
B1 = y 4 ,


4−r 4
C1 = p z ,
or

4−r 4

A1 = p z ,
(8)
B1 = y 4 ,


4
C1 = x ,
where x, y, pz are pairwise coprime.
If A1 , B1 and C1 satisfy (7), then
x4 + y 4 = p4−r z 4 .
(9)
But gcd(y, p) = 1, we have integer s 6≡ 0 (mod p) such that sy ≡ 1 (mod p). By
(9) we deduce that
(xs)4 ≡ −1 (mod p),
which implies that −1 is a square modulo p, this can only hold for p = 2 and p ≡ 1
(mod 4), in contradiction with p ≡ 3 (mod 8).
If A1 , B1 and C1 satisfy (8), then
x4 − y 4 = p4−r z 4 .
(10)
6
TIANXIN CAI, DEYI CHEN AND YONG ZHANG
When r = 2, we have x4 − y 4 = (pz 2 )2 , but it is well known that the equation
X 4 − Y 4 = Z 2 has no nonzero integer solutions. So r = 1 or r = 3 and (10) can be
changed into
x4 − y 4 = p(pz 2 )2
(11)
or
x4 − y 4 = p(z 2 )2
(12)
respectively. By Lemma 1 and Lemma 2, we obtain that (11) and (12) have no
positive integer solutions when p ≡ 3 (mod 8).
Proof of Theorem 4. In view of Lemma 4, we only need to discuss k ≡
6 0
(mod 5). In view of gcd(A, B, C) = pk , (4) can be changed into
(
A1 + B1 = C1
(13)
p3k A1 B1 C1 = D5 .
where A1 = pAk , B1 = pBk and C1 = pCk are pairwise coprime. But k 6≡ 0 (mod 5),
we deduce that only one of A1 , B1 , C1 congruent to 0 modulo p. Let 3k ≡ r
(mod 5), then 1 ≤ r ≤ 4. In view of (13) and reorder A1 , B1 , C1 if necessary, we
obtain

A1 = x5

(14)
B1 = y 5 ,


5−r 5
C1 = p z ,
where x, y and pz are pairwise coprime. From (13) and (14) we deduce that
x5 + y 5 = p5−r z 5 .
(15)
But p is an odd prime and p 6≡ 1 (mod 10), by Lemma 3, (15) has no positive
integer solutions.
k
k
+y
> 0 for any odd
Proof of Theorem 5. First of all, we prove that f (x, y) = xx+y
2
positive integer k, where (x, y) ∈ R , x + y 6= 0. It is obvious that f (x, y) > 0 when
x+y
xy ≥ 0. So we only need to prove the case xy < 0. If k = 1, f (x, y) = x+y
= 1 > 0.
Suppose
xk +y k
x+y
> 0 where x + y 6= 0. Because xy < 0 and x + y 6= 0, we obtain
xk+2 + y k+2
xk + y k
=
(−xy) + xk+1 + y k+1 > 0.
x+y
x+y
k
k
+y
> 0 for any odd
By mathematical induction, we deduce that f (x, y) = xx+y
2
positive integer k, where (x, y) ∈ R , x + y 6= 0. By condition of Theorem 5, we
have a prime p such that
n−1
X
an + b n
n−1
n−2
n−3 2
n−1
0<p=
an−1−j (−b)j . (16)
=a
−a
b+a
b − ···+ b
=
a+b
j=0
Since n ≥ 3, we deduce from (16) that
gcd(p, a) = gcd(p, b) = gcd(a, b) = 1.
(17)
gcd(a, a + b) = gcd(b, a + b) = 1.
(18)
So
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
7
Let

rn−1
+tn n

a
A = p 3
rn−1
(19)
B = p 3 +tn bn

rn+2

+tn
(a + b)
C=p 3
where r is defined in the condition of Theorem 5 and integer t ≥ 0.
By (17), (18), (19) and a + b = mn , we obtain that A, B and C satisfy (1) where
D = pr+3t abm and
gcd(A, B, C) = pk ,
where k = rn−1
+ tn ≡ rn−1
(mod n).
3
3
Finally, we want to prove that p ≡ 1 (mod 2n). We only need to prove that p ≡ 1
(mod n) because both p and n are odd primes.
n
+bn
, n ≥ 3 is prime, by Fermat’s little Theorem, we have
Since p = a a+b
p(a + b) = an + bn ≡ a + b
(mod n).
Now we only need to prove that a + b 6≡ 0 (mod n). Suppose a + b ≡ 0 (mod n),
then −b ≡ a (mod n) and
an + b n
p=
= an−1 − an−2 b + an−3 b2 − · · · + bn−1 ≡ nan−1 ≡ 0 (mod n).
a+b
But p and n are two primes, so we deduce that p = n.
If ab < 0, We may assume that a > 0, b < 0, then
an + b n
= an−1 − an−2 b + an−3 b2 − · · · + bn−1 > 1| + 1 +{z· · · + 1} = n = p.
p=
a+b
n
contradiction.
n
+bn
is prime, so a 6= b and
If ab ≥ 0, we may assume that a ≥ 0, b ≥ 0, but p = a a+b
ab 6= 0. Reorder a, b (if necessary), we may assume that a ≥ 1, b ≥ 2.
n
1+bp
p
If a = 1,b ≥ 2, then p = 1+b
1+b = 1+b . Let f (b) = (b + 1) − p(b + 1), then
so
f ′ (b) = pbp−1 − p = p(bp−1 − 1) > 0,
f (b) ≥ f (2) = (2p + 1) − 3p > 0
p
for odd prime p, from which we deduce that 1+b
1+b > p. Contradiction.
n
p
+bn
+bp
If a ≥ 2, b ≥ 2, then p = a a+b
= aa+b
> pa+pb
a+b = p, once again a contradiction!
Finally, we list some primes p satisfying the condition of Theorem 5 when
n = 5, 7.
8
TIANXIN CAI, DEYI CHEN AND YONG ZHANG
Table 1. Primes p < 107 which satisfy the condition of Theorem
5 when n = 5.
p
31
211
4651
61051
132661
202981
371281
723901
a
b
m
2 −1 1
3 −2 1
6 −5 1
11 −10 1
11 21
2
9
23
2
17 −16 1
20 −19 1
1641301
1803001
2861461
4329151
4925281
5754901
7086451
7944301
8782981
35
25
28
31
32
45
35
36
49
−3
−24
−27
−30
−31
−13
−34
−35
−17
2
1
1
1
1
2
1
1
2
Table 2. Primes p < 1011 which satisfy the condition of Theorem
5 when n = 7.
p
a
b
m
127
2 −1 1
14197
4 −3 1
543607
7 −6 1
1273609
8 −7 1
2685817
9 −8 1
5217031 10 −9 1
16344637 12 −11 1
141903217 17 −16 1
1928294551
8258704609
14024867221
22815424087
30914273881
77617224511
91154730577
98201826199
26
33
36
39
41
59
49
55
−25
−32
−35
−38
−40
69
−48
73
1
1
1
1
1
2
1
2
4. New Fermat equation in quadratic fields
It’s well-known that Fermat’s equation xn +y n = z n has only the trivial solutions
in integers when n ≥ 3. Therefore, it is an interest problem that whether Fermat’s
equation has non-trivial solutions in algebraic number fields. There are numerus
papers on this problem and we can refer to [1, 6] and the references there. For the
case n = 3, it was solved almost completely. In 1915, W. Burnside [1] proved that
in quadratic field Fermat’s equation x3 + y 3 = z 3 has solutions of the form
p

x
=
−3
+
−3(1 + 4k 3 ),


p
y = −3 − −3(1 + 4k 3 ),


z = 6k,
where k is a rational number not equal to 0 and√−1. While k = 0, Fermat’s equation
x3 + y 3 = z 3 has no non-trivial solutions in Q −3.
In 2013, M. Jones and J. Rouse [6] gave necessary and sufficient conditions on a
3
3
3
square-free
√ integer t such that x + y = z has a nontrivial solution in quadratic
fields Q( t), under the Birch and Swinnerton-Dyer conjecture.
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
9
Now we consider the new Fermat equation (1) for n√ = 3 in quadratic fields
√
√
Q( t). Assume that the solution in quadratic fields Q( t) has the form a + b t
with ab 6= 0. We have the following theorem.
Theorem 6. For any square-free integer t 6= 0, 1 such that the elliptic curve
tu2 = 1 + 4k 3
has a √
nonzero rational solution (u, k), then (1) has infinitely many solutions (A, B, C, D)
in Q( t) for n = 3.
√
√
√
Proof of Theorem 6. Let A = a + b t, B = c + d t, D = e + f t. When n = 3,
we get from (1) that
a2 c + ac2 + 2adtb + ad2 t + b2 tc + 2btcd + (2acb + 2acd + a2 d
√
√
+ bc2 + tb2 d + tbd2 ) t = e3 + 3ef 2 t + f (3e2 + f 2 t) t.
Then
(
a2 c + ac2 + 2adtb + ad2 t + b2 tc + 2btcd = e3 + 3ef 2 t,
2acb + 2acd + a2 d + bc2 + tb2 d + tbd2 = f (3e2 + f 2 t).
Solving the above two equations, we have
a2 c + ac2 − e3
2acb + 2acd + a2 d + bc2 − 3f e2
=−
.
2
+
+ 2bcd + 2adb − 3ef
b2 d + bd2 − f 3
Taking e = kc, f = kd, from the formula of t, we have
t=−
ad2
b2 c
(ad + 2ab + cb − 2ck 3 d)(d2 a2 + c2 b2 + 2cbda + 2c2 db + 2cd2 a − 4c2 d2 k 3 ) = 0.
Let us consider ad + 2ab + cb − 2ck 3 d = 0, then
b(c + 2a)
.
d=−
a − 2ck 3
Hence, we get
(a − 2ck 3 )2
t=
.
(1 + 4k 3 )b2
Put a − 2ck 3 = (1 + 4k 3 )b, then
t = 1 + 4k 3 .
Therefore, for n = 3, (1) has solutions
p


A = (1 + 4k 3 )a + (a − 2k 3 c) 1 + 4k 3 ,


p


 B = (1 + 4k 3 )c − (2a + c) 1 + 4k 3 ,
p
3
3


1 + 4k 3 ,
C
=
(1
+
4k
)(a
+
c)
−
(a
+
(2k
+
1)c)


p


D = (1 + 4k 3 )kc − k(2a + c) 1 + 4k 3 ,
where a, c and k are non-zero rational numbers.
Let tu2 = 1 + 4k 3 , where t is a square-free integer and t 6= 0, 1. Then for n = 3
the new Fermat equation (1) has solutions
√

3
A
=
uta
+
(a
−
2k
c)
t,



√

 B = utc − (2a + c) t,
√

C = ut(a + c) − (a + (2k 3 + 1)c) t,



√

D = utkc − k(2a + c) t.
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TIANXIN CAI, DEYI CHEN AND YONG ZHANG
This completes the proof of Theorem 6.
As an example, taking k = −1, then (1) has solutions
√

A = −3a + (a + 2c) −3,




 B = −3c − (2a + c)√−3,
√

C
=
−3(a
+
c)
−
(a
−
c)
−3,



√

D = 3c + (2a + c) −3,
where a, c are non-zero rational numbers.
To get Burnside’s solutions for a given t, we need to consider the following elliptic
curve
tu2 = −3(1 + 4k 3 ).
By some calculations, we find that the elliptic curves tu2 = 1 + 4k 3 and tu2 =
−3(1 + 4k 3 ) have the same j-invariant, so they are isomorphic and have the same
rank. If they have the rank greater than 1, then there are infinitely many rational
solutions (u, k) for both of these two elliptic curves. If they have the rank zero, we
can’t distinguish the torsion points on them, they might have or not have non-zero
rational solutions. For −50 ≤ t ≤ 50 and t is square-free, we find no other t as the
above example t = −3. So we may ask the following question.
Question: Are there other t 6= −3 such that for n = 3 the new Fermat equation
(1) has non-trivial solutions but the Fermat equation x3 + y 3 = z 3 hasn’t?
References
1. W. Burnside, On the rational solutions of the equation x3 + y 3 + z 3 = 0 in quadratic fields,
Proc. London Math. Soc. 14 (1915), 1-4.
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2333-2341.
3. H. Cohen, Number Theory, Volume I: Tools and diophantine equations, Graduate Texts in
Mathematics, vol. 239, Springer-Verlag, New York, 2007.
4. H. Darmon and A. Granville, On the equations z m = F (x, y) and Axp + By q = cZ r , Bull.
London Math. Soc. 27(1995), 513-543.
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(2004), no. 2, 291-302.
6. M. Jones and J. Rouse, Solutions of the cubic Fermat equation in quadratic fields, Int. J.
Number Theory 9 (2013), 1579-1591.
7. V. A. Lebesgue, Théorémes nouveaux sur l’équation indéterminée x5 + y 5 = az 5 , J. Math.
Pures Appl. 8 (1843) 49-70.
8. R. D. Mauldin, A Generalization of Fermat’s Last Theorem: The Beal Conjecture and Prize
Problem. Notices of the AMS 44 (11) (1997): 1436-1439.
9. S. Mochizuki, Inter-Universal Teichmüller Theory IV: Log-Volume Computations and SetTheoretic Foundations. http://www.kurims.kyoto-u.ac.jp/˜motizuki/papers-english.html
10. C. Pomerance, Computational Number Theory, Leader, Imre, The Princeton Companion to
Mathematics, Princeton University Press, pp. 361-362, June 2008.
11. R. Taylor and A. Wiles, Ring-theoretic properties of certain Hecke algebras. Ann. of Math.
(1995), 553-572.
12. J. B. Tunnell, A classical Diophantine problem and modular forms of weight 3/2. Invent.
Math. 72 (2) (1983), 323-334.
13. A. Wiles, Modular elliptic curves and Fermat’s last theorem. Ann. of Math. (1995), 443-551.
14. The Beal Prize, AMS, http://www.ams.org/profession/prizes-awards/ams-supported/bealprize
A NEW GENERALIZATION OF FERMAT’S LAST THEOREM
Department of Mathematics, Zhejiang University, Hangzhou, 310027, China
Email address: [email protected]
Department of Mathematics, Zhejiang University, Hangzhou, 310027, China
Email address: [email protected]
Department of Mathematics, Zhejiang University, Hangzhou, 310027, China
Email address: [email protected]
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